A293207 Lexicographically earliest sequence of positive terms such that the function f defined by f(n) = Sum_{k=1..n} (i^k * a(k)) for any n >= 0 is injective (where i denotes the imaginary unit), and a(n) != a(n+1) and a(n) != a(n+2) for any n > 0.

1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 4, 1, 2, 4, 1, 2, 3, 1, 4, 5, 1, 7, 2, 1, 4, 2, 1, 4, 5, 1, 2, 3, 1, 2, 6, 3, 1, 4, 2, 1, 9, 2, 1, 3, 2, 1, 11, 2, 1, 6, 2, 1, 3, 2, 1, 6, 4, 7, 1, 3, 2, 1, 3, 9, 1, 2, 3, 1, 4, 7, 5, 1, 2, 4, 1, 11, 4, 10, 1, 9, 2, 6, 7, 1, 9

Offset: 1

Author

Rémy Sigrist, Oct 02 2017

Keywords

• nonn

Comments

See A293208 for the real part of f(n).

See A293209 for the imaginary part of f(n).

For any m > 0, let b_m be the lexicographically earliest sequence of positive terms such that the function f_m defined by f_m(n) = Sum_{k=1..n} (i^k * b_m(k)) for any n >= 0 is injective, and #{b_m(n), ..., b_m(n+m)} = m+1 for any n > 0:

- in particular, b_2 = a (this sequence) and f_2 = f,

- the representation of f shows an apparently chaotic initial phase followed by the emergence of a regular oscillating escape (see representations in Links section),

- the cases m=8 and m=12 have similarities with Langton's ant: the representation of f_m shows an apparently chaotic initial phase followed by the emergence of a regular escape (see representations in Links section),

- for the cases m=3, m=7, m=11 and m=15: b_m is m+1 periodic and b_m(n) = n for any n <= m+1,

- for the cases m=4, m=5, m=6, m=9, m=10, m=13, m=14 and m=16: the representation of f_m shows an apparently chaotic initial phase; it is unknown whether a regular escape emerges (see representation in Links section).

More informally, the sequence can be obtained with the following procedure:

- start at the origin, looking to the north,

- repeatedly: jump forward to the nearest non-visited square (provided that the jump length is distinct from the two previous jump lengths) and turn 90 degrees to the left,

- a(n) = length of n-th jump and f(n-1) = position before n-th jump as a Gaussian integer.

Examples

			f(0) = 0
i^1 = i.
f(0) + 1*i has not yet been visited; hence a(1) = 1 and f(1) = i.
i^2 = -1.
f(1) + 1*-1 has not yet been visited, but a(1) = 1.
f(1) + 2*-1 has not yet been visited; hence a(2) = 2 and f(2) = -2 + i.
i^3 = -i.
f(2) + 1*-i has not yet been visited, but a(1) = 1.
f(2) + 2*-i has not yet been visited, but a(2) = 2.
f(2) + 3*-i has not yet been visited; hence a(3) = 3 and f(3) = -2 - 2*i.
i^4 = 1.
f(3) + 1*1 has not yet been visited; hence a(4) = 1 and f(4) = -1 - 2*i.
i^5 = i.
f(4) + 1*i has not yet been visited, but a(4) = 1.
f(4) + 2*i has not yet been visited; hence a(5) = 2 and f(5) = -1.
i^6 = -1.
f(5) + 1*-1 has not yet been visited, but a(4) = 1.
f(5) + 2*-1 has not yet been visited, but a(5) = 2.
f(5) + 3*-1 has not yet been visited; hence a(6) = 3 and f(6) = -4.
i^7 = -i.
f(6) + 1*-i has not yet been visited; hence a(7) = 1 and f(7) = -4 - i.
i^8 = 1.
f(7) + 1*1 has not yet been visited, but a(7) = 1.
f(7) + 2*1 has not yet been visited; hence a(8) = 2 and f(8) = -2 - i.
i^9 = i.
f(8) + 1*i has not yet been visited, but a(7) = 1.
f(8) + 2*i has not yet been visited, but a(8) = 2.
f(8) + 3*i has not yet been visited; hence a(9) = 3 and f(9) = -2 + 2*i.
i^10 = -1.
f(9) + 1*-1 has not yet been visited; hence a(10) = 1 and f(10) = -3 + 2*i.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..60000
• Rémy Sigrist, Representation of f(n) for n=0..60000
• Rémy Sigrist, Representation of f_8(n) for n=0..121346
• Rémy Sigrist, Representation of f_12(n) for n=0..4463502
• Rémy Sigrist, Representation of f_6(n) for n=0..10000000
• Rémy Sigrist, PARI program for A293207
• Wikipedia, Langton's ant

Crossrefs

See A293208, A293209.

Programs

PARI

See Links section.

Formula

a(56948 + 8*k + i) = (1-k) * a(56948 + i) + k * a(56948 + i + 8) for any k >= 0 and i in 0..7.